Math 523H Homework #3
Sections 1.4, 2.1
1.4 #1.
Give 5 examples which show that P implies Q does not necessarily mean
that Q implies P.
•
If
x
∈
R
and
x >
0, then
x
2
>
0. However, if
x
2
>
0,
x
is not necessarily greater than 0.
•
If
x
∈
R
and
x
≥
0, then

x
 ≥
0. But if

x
 ≥
0, then
x
is not necessarily greater than or
equal to 0.
•
If
x
and
y
are even integers, then
x
+
y
is an even integer. But if
x
+
y
is an even integer,
then
x
and
y
are not necessarily even.
•
If
a
and
b
are rational numbers, then
a
b
is a rational number. But if
a
b
is a rational number,
then
a
and
b
are not necessarily rational.
•
If
x
and
y
are both even integers, then
xy
is an even integer. But if
xy
is an even integer,
then it is not necessarily true that both
x
and
y
are even.
1.4 #4. Suppose that
0
< a < b
. Prove that:
(a)
a <
√
ab < b
.
Proof.
a < b
. Thus,
a
·
a < a
·
b
, or
a
2
< ab
. Since
a >
0 and
b >
0,
ab >
0 so
√
ab
exists in
R
. Thus, we can are able to take the square root of both sides, and we get
√
a
2
<
√
ab
, or
a <
√
ab
. Since
a < b
, we also have
a
·
b < b
·
b
, or
ab < b
2
. So
√
ab <
√
b
2
, or
√
ab < b
. By
transitivity,
a <
√
ab < b
.
(b)
√
ab
≤
1
2
(
a
+
b
).
Proof.
For any
a, b
∈
R
, 0
≤
(
a

b
)
2
, or 0
≤
a
2

2
ab
+
b
2
.
By (O4) of section 1.1,
0 + 4
ab
≤
(
a
2

2
ab
+
b
2
) + 4
ab
, or 4
ab
≤
a
2
+ 2
ab
+
b
2
.
But
a
2
+ 2
ab
+
b
2
= (
a
+
b
)
2
,
so 4
ab
≤
(
a
+
b
)
2
. Since
a >
0 and
b >
0, 4
ab >
0 and thus
√
4
ab
exists in
R
. Thus, we
can take the square root of both sides, and
√
4
ab
≤
p
(
a
+
b
)
2
. This may be simplified to
2
√
ab < a
+
b
. Since
1
2
>
0, we can use (O5) of section 1.1, and we get
1
2
(2
√
ab
)
≤
1
2
(
a
+
b
),
or
√
ab
≤
1
2
(
a
+
b
).
1.4 #7. Prove that for any
n
natural numbers,
x
1
, x
2
, ..., x
n
,

x
1
+
x
2
+
...
+
x
n
 ≤ 
x
1

+

x
2

+
...
+

x
n

.
Proof.
For
n
= 1,

x
1

=

x
1

, so the inequality
≤
holds for
n
= 1. Assume the above statement is
also true for the first
n
natural numbers.

x
1
+
x
2
+
...
+
x
n
+
x
n
+1

=

(
x
1
+
x
2
+
...
+
x
n
)+
x
n
+1

,
so by the Triangle Inequality,

(
x
1
+
x
2
+
...
+
x
n
)+
x
n
+1
 ≤ 
x
1
+
x
2
+
...
+
x
n

+

x
n
+1

. Because we
assumed that the statement above was true for the first
n
natural numbers,

x
1
+
x
2
+
...
+
x
n
 ≤

x
1

+

x
2

+
...
+

x
n

. Therefore,

x
1
+
x
2
+
...
+
x
n

+

x
n
+1
 ≤ 
x
1

+

x
2

+
...
+

x
n

+

x
n
+1

. By
transitivity,

x
1
+
x
2
+
...
+
x
n
+
x
n
+1
 ≤ 
x
1

+

x
2

+
...
+

x
n

+

x
n
+1

, so the statement is also
true for the first
n
+ 1 natural numbers. By induction, we have proven that

x
1
+
x
2
+
...
+
x
n
 ≤

x
1

+

x
2

+
...
+

x
n

for any
n
∈
N
.
1
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1.4 #8. Prove that for all positive integers
n
,
1
3
+ 2
3
+
...
+
n
3
= (1 + 2 + 3 +
...
+
n
)
2
.
Proof.
For
n
=1, we have 1
3
= 1 = 1
2
. So the statement holds for
n
= 1. Now assume the above
statement is true for the first
n
natural numbers, that is 1
3
+ 2
3
+
...
+
n
3
= (1 + 2 + 3 +
...
+
n
)
2
.
By (O4), 1
3
+ 2
3
+
...
+
n
3
+ (
n
+ 1)
3
= (1 + 2 + 3 +
...
+
n
)
2
+ (
n
+ 1)
3
. But by Proposition
1.4.3, 1 + 2 +
...
+
n
=
n
(
n
+1)
2
, so we have 1
3
+
...
+
n
+ (
n
+ 1)
3
= (
n
(
n
+1)
2
)
2
+ (
n
+ 1)
3
=
n
2
(
n
+1)
2
4
+
4(
n
+1)
3
4
=
(
n
+1)
2
(
n
2
+4
n
+4)
4
=
(
n
+1)
2
(
n
+2)
2
4
= (
(
n
+1)(
n
+2)
2
)
2
= (1+2+
...
+
n
+(
n
+1))
2
.
Therefore, 1
3
+
...
+
n
3
+(
n
+1)
3
= (1+2+
...
+
n
+(
n
+1))
2
, so the statement is also true for the first
n
+1 natural numbers. By induction, we have proven that 1
3
+2
3
+
...
+
n
3
= (1+2+3+
...
+
n
)
2
.
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 Fall '10
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 Math, Natural Numbers, lim, Natural number

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